integral from 0 to 1 of sqrt(4x-x^2)

Lösung

\int_{0}^{1} 4 x - x^{2} d x

2 (\frac{π}{3} - \frac{3}{4})

Dezimale

1.22836 \dots

Schritte zur Lösung

\int_{0}^{1} 4 x - x^{2} d x

Vervollständige das Quadrat

4 x - x^{2} : - (x - 2)^{2} + 4

= \int_{0}^{1} - (x - 2)^{2} + 4 d x

Wende U-Substitution an

= \int_{- 2}^{- 1} - u^{2} + 4 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{- \frac{π}{6}} 4 cos^{2} (v) d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{- \frac{π}{2}}^{- \frac{π}{6}} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \cdot \int_{- \frac{π}{2}}^{- \frac{π}{6}} \frac{1 + cos ( 2 v )}{2} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{- \frac{π}{6}} 1 + cos (2 v) d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{- \frac{π}{6}} 1 d v + \int_{- \frac{π}{2}}^{- \frac{π}{6}} cos (2 v) d v)

\int_{- \frac{π}{2}}^{- \frac{π}{6}} 1 d v = \frac{π}{3}

\int_{- \frac{π}{2}}^{- \frac{π}{6}} cos (2 v) d v = - \frac{3}{4}

= 4 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Vereinfache

4 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 2 (\frac{π}{3} - \frac{3}{4})

= 2 (\frac{π}{3} - \frac{3}{4})

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